According to Lutkepohl (2005) autocovariance matrix for VAR(1) case is given by \begin{equation} \begin{aligned} \Gamma_y(h) &= E(y_t - \mu)(y_{t-h} - \mu)' \\ &= \lim_{n\to\infty}\sum_{i=0}^{n}\sum_{j=0}^{n}A_1^iE(u_{t-i}u'_{t-h-j})(A_1^j)' \\ &= \lim_{n\to\infty}\sum_{i=0}^{n}A_1^{h+i}\Sigma_u(A_1^i)' \\ &= \sum_{i=0}^{\infty}A_1^{h+i}\Sigma_u(A_1^i)' \end{aligned} \end{equation} since $E(u_tu_s') = 0$ for $s\neq t$ and $E(u_tu_t') = \Sigma_u \;\; \forall t$.

I have troubles with getting from second to third line. I understand that since $E(u_tu_s') = 0$ for $s\neq t$, the term $\sum_{i=0}^{n}\sum_{j=0}^{n}A_1^iE(u_{t-i}u'_{t-h-j})(A_1^j)'$ is zero unless $i = h+j$ (and hence $j = i-h$). So I remove summation over $j$ and set $j = i - h$. I get expression $$\sum_{i=0}^{\infty}A_1^i\Sigma_u(A_1^{i-h})'.$$

How do I get to the expression in the book?

  • $\begingroup$ The autocovariance function is an even function of $h$. Try removing summation over $i$, setting $i=h+j$, to get $$\Gamma_y(h) = \sum_{j=0}^{\infty}A_1^{h+j}\Sigma_u(A_1^j)'.$$ Then, replace $j$ by $i$ $\endgroup$ – Dilip Sarwate Mar 2 '17 at 14:34
  • $\begingroup$ @Dilip Sarwate, thanks for your comment, that worked. $\endgroup$ – tosik Mar 2 '17 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.